This Exam P sample reference tests Uniform Distribution. The no-deductible mean identifies the uniform support as [1,19]. A deductible of 4 leaves a positive excess over a 15-unit interval; integrating that excess against density 1/18 gives 225/36=6.25, so choice B is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis subtracts the deductible directly from the original mean, 10-4=6, and allows negative payments for losses below 4.
CThis is the mean excess conditional on a payment, (19-4)/2=7.5, and omits the probability that a payment occurs.
DThis scales the original mean by the fraction (19-4)/19, treating the support as if it began at zero instead of integrating the excess.
EThis multiplies the original mean 10 by P(X>4)=15/18 but does not subtract the deductible from paid losses.
Original practice · fully worked
Original variant: research-vessel repair threshold
A research vessel's repair-cost index X is uniformly distributed from 0 to 24. A reserve pays the amount of X above an unknown activation threshold d and pays zero otherwise. The expected reserve payment is 3. Determine d.
A 3
B 9
C 12
D 18
E 21
Variant answer in brief
For a uniform index on [0,24], the expected positive excess over d is (24-d)²⁄⁴⁸. Equating this to 3 gives 24-d=12 and hence d=12, so choice C is correct.
Setup
Setup
Represent the reserve payment as the positive part above the unknown threshold.
X∼Uniform(0,24)
R=(X−d)+
Model
Model
Integrate the excess payment using the constant uniform density.
E[R]=∫d24(x−d)241dx
E[R]=48(24−d)2
Compute
Compute
Set the expected payment equal to its specified value and retain the threshold within the support.
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